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# a,b,c are three different numbers.None of the numbers equals the

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Current Student
Joined: 12 Aug 2015
Posts: 2617
Schools: Boston U '20 (M)
GRE 1: Q169 V154
a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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Updated on: 19 Apr 2017, 13:36
6
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:11) correct 31% (02:24) wrong based on 64 sessions

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a,b,c are three different numbers.None of the numbers equals the average of the other two.
If $$\frac{x}{{a+b-2c}}$$ = $$\frac{y}{{b+c-2a}}$$ = $$\frac{z}{{c+a-2b}}$$ , then what is the value of $$x+y+z$$ ?

A)-1
B)5
C)3
D)1
E)0

Source-> NOVA.

_________________

Originally posted by stonecold on 19 Apr 2017, 13:32.
Last edited by Bunuel on 19 Apr 2017, 13:36, edited 1 time in total.
Renamed the topic.
Senior Manager
Joined: 24 Apr 2016
Posts: 331
Re: a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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19 Apr 2017, 13:36
3
1
Let

x/(a+b-2c) = y/(b+c-2a) = z/(c+a-2b) = k

which means

x = (a+b-2c)k
y = (b+c-2a)k
z = (c+a-2b)k

x+y+z = k*[a+b-2c+b+c-2a+c+a-2b] = k*0 = 0

Manager
Joined: 12 Sep 2017
Posts: 247
a,b,c are three different numbers.None of the numbers equals the  [#permalink]

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03 Mar 2019, 19:18
stonecold wrote:
a,b,c are three different numbers.None of the numbers equals the average of the other two.
If $$\frac{x}{{a+b-2c}}$$ = $$\frac{y}{{b+c-2a}}$$ = $$\frac{z}{{c+a-2b}}$$ , then what is the value of $$x+y+z$$ ?

A)-1
B)5
C)3
D)1
E)0

Source-> NOVA.

Hello stonecold !

Could you please provide the explanation?

Kind regards!
a,b,c are three different numbers.None of the numbers equals the   [#permalink] 03 Mar 2019, 19:18
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# a,b,c are three different numbers.None of the numbers equals the

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